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Precalculus : Pre-Calculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Find The Roots Of Complex Numbers

Solve for z^3-z^2+z=0 (there may be more than one solution).

Possible Answers:

$0, 1, \frac{1}{2}$

$0, \frac{1}{2}\pm\frac{\sqrt{3}}{2}$

$0, \frac{1}{2}\pm\frac{\sqrt{3i}}{2}$

$0, \frac{1}{2}\pm\sqrt{3i}$

Correct answer:

$0, \frac{1}{2}\pm\frac{\sqrt{3i}}{2}$

Explanation:

To solve for the roots, just set equal to zero and solve for z using the quadratic formula ( $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ) : $z^3-z^2+z\rightarrow z(z^2-z+1)$ and now setting both and z^2-z+1 equal to zero we end up with the answers z=0 and $z=\frac{1\pm 3i}{2}$

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Example Question #1 : Find The Roots Of Complex Numbers

Compute (3+3i)^5

Possible Answers:

972

972i

972+972i

-972-972i

972-972i

Correct answer:

-972-972i

Explanation:

To solve this question, you must first derive a few values and convert the equation into exponential form: $re^{\pi*z}$ : $r=\sqrt{9+9}=3\sqrt{2}\rightarrow tan(\theta)=\frac{3}{3}=1\rightarrow Arg(z)=\frac{\pi}{4}$

Now plug back into the original equation and solve: $(3+3i)^5=(3\sqrt{2})^5\cdot e^{\frac{i5\pi}{4}=972\sqrt{2}(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4}))=972\sqrt{2}(\frac {-\sqrt{2}{2}}-\frac ({\sqrt{2}\cdot i}{2})=-972-972i$

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Example Question #2 : Find The Roots Of Complex Numbers

Determine the length of 2i^2-1

Possible Answers:

Correct answer:

Explanation:

i^2=-1 , so $2(-1)-1=-3\rightarrow |-3|=3$

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Example Question #3 : Find The Roots Of Complex Numbers

Solve for all possible solutions to the quadratic expression: m^2-6m+12=0

Possible Answers:

$3\pm3i$

$3\pm\sqrt{3}i$

$3\pm\sqrt{3}$

$3\pm3$

Correct answer:

$3\pm\sqrt{3}i$

Explanation:

Solve for complex values of m using the aforementioned quadratic formula:

$\frac{6\pm\sqrt{36-(4)(1)(12)}}{2}$

$=\frac{6\pm\sqrt{-12}}{2}$

$=\frac{6\pm2i\sqrt{3}}{2}$

$=3\pm \sqrt{3}i$

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Example Question #4 : Find The Roots Of Complex Numbers

Which of the following lists all possible solutions to the quadratic expression: $m^{2}-6m+12=0$

Possible Answers:

$3\pm \sqrt{3}$

$3\pm 3i$

$3\pm 3$

$3\pm \sqrt{3i}$

Correct answer:

$3\pm \sqrt{3i}$

Explanation:

Solve for complex values of using the quadratic formula: $6\pm \sqrt{36-(4)(1)(12)}\rightarrow \frac{6\pm \sqrt{-12}}{2}=3\pm \sqrt{3i}$

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Example Question #10 : Find The Roots Of Complex Numbers

Determine the length of $2i^{2}-1$ .

Possible Answers:

Correct answer:

Explanation:

To begin, we must recall that $i^{2}=-1$ . Plug this in to get 2i^2-1=2(-1)-1=-2-1=-3 . Length must be a positive value, so we'll take the absolute value: $\left | -3 \right |=3$ . Therefore the length is 3.

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Example Question #11 : Find The Roots Of Complex Numbers

Solve for $z^{3}-z^{2}+z=0$ (there may be more than one solution).

Possible Answers:

$0,\frac{1}{2},\pm \frac{\sqrt{3i}}{2}$

$0,\frac{1}{2},\pm \sqrt{3i}$

$0,1,\frac{1}{2}$

$0,\frac{1}{2},\pm \frac{\sqrt{3}}{2}$

Correct answer:

$0,\frac{1}{2},\pm \frac{\sqrt{3i}}{2}$

Explanation:

To solve for the roots, just set equal to zero and solve for z using the quadratic formula, which is $\frac{-b\pm \sqrt{b^{2}4ac}}{2a}$

$z^{3}-z^{2}+z\rightarrow z\left ( z^{2}-z+1 \right )$ and now setting both and $z^{2}-z+1$ equal to zero we end up with the answers z=0 and $z=\frac{1\pm 3i}{2}$ and so the correct answer is $0,\frac{1}{2},\pm \frac{2\sqrt{3i}}{2}$ .

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Example Question #12 : Find The Roots Of Complex Numbers

Solve for all possible solutions to the quadratic expression: $m^{2}-6m+12=0$

Possible Answers:

$3\pm \sqrt{3}$

$3\pm 3$

$3\pm 3i$

$3\pm \sqrt{3i}$

Correct answer:

$3\pm \sqrt{3i}$

Explanation:

Solve for complex values of using the quadratic formula: $6\pm\sqrt{36-(4)(1)(12)} = \frac{6\pm\sqrt{-12}}{2}=\frac{6\pm2\sqrt{3}i}{2}=3\pm\sqrt{3}i$ .

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Example Question #43 : Polar Coordinates And Complex Numbers

Solve for $z^{3}-z^{2}+z=0$ (there may be more than one solution).

Possible Answers:

$0,\frac{1}{2},\pm \sqrt{3i}$

$0,1,\frac{1}{2}$

$0,\frac{1}{2},\pm \frac{\sqrt{3i}}{2}$

$0,\frac{1}{2},\pm \frac{\sqrt{3}}{2}$

Correct answer:

$0,\frac{1}{2},\pm \frac{\sqrt{3i}}{2}$

Explanation:

To solve for the roots, just set equal to zero and solve for using the quadratic formula ( $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ ): $z^{3}-z^{2}+z\rightarrow z(z^{2}-z+1)$ and now setting both and $z^{2}-z+1$ equal to zero we end up with the answers z=0 and $z=\frac{1\pm 3i}{2}$ .

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Example Question #1 : Polar Form Of Complex Numbers

Convert the following to rectangular form:

2(cos(30)+ isin(30))

Possible Answers:

$\sqrt{3}+i$

$2\sqrt{3}+2i$

$\sqrt{3}+2i$

$\sqrt{3}-1$

$2\sqrt{3}+1$

Correct answer:

$\sqrt{3}+i$

Explanation:

Distribute the coefficient 2, and evaluate each term:

$\\2(cos(30)+ isin(30))\\ \\= 2cos(30)+i(2sin(30))\\ \\=\sqrt{3}+i$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #2 : Find The Roots Of Complex Numbers

Example Question #3 : Find The Roots Of Complex Numbers

Example Question #4 : Find The Roots Of Complex Numbers

Example Question #10 : Find The Roots Of Complex Numbers

Example Question #11 : Find The Roots Of Complex Numbers

Example Question #12 : Find The Roots Of Complex Numbers

Example Question #43 : Polar Coordinates And Complex Numbers

Example Question #1 : Polar Form Of Complex Numbers

All Precalculus Resources

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